Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Refer to exercises 9, 10 of chapter 3 in Lang's algebra, page 167.

In particular, let $A$ be a commutative ring, $p$ a prime ideal and $M, N$ $A$-modules. Then $M_p, N_p$ are the localized $A_p$ modules. Let $f: M_p \rightarrow N_p$ be a homomorphism of $A_p$ modules. How can we lift this homomorphism to an $A$-homomorphism $M \rightarrow N$ of the initial modules?

share|cite|improve this question
Dear Manos, if you refer to an exercise in Lang, you should copy it in your question and not force users to try to get hold of that book. – Georges Elencwajg Oct 31 '11 at 9:42
up vote 3 down vote accepted

You are starting with a sequence of $A$-modules $$ 0 \to M' \stackrel f\to M \stackrel g\to M'' \to 0. $$ You want to show that if for each prime $\mathfrak{p}$ of $A$ the induced sequence of $A_\mathfrak{p}$-modules $$ 0 \to M'_\mathfrak{p} \stackrel{f_\mathfrak{p}}\longrightarrow M_\mathfrak{p} \stackrel{g_\mathfrak{p}}\longrightarrow M''_\mathfrak{p} \to 0 $$ is exact, then the original sequence is exact. Here the map $f_\mathfrak{p}$, for example, just sends $x/s$ to $f(x)/s$. You don't need to perform any sort of "lifting" on $f_\mathfrak{p}$, because you already have $f$. One way of solving the problem is to find ways to apply the fact that if $N_\mathfrak{p} = 0$ for all primes $\mathfrak{p}$, then $N = 0$. [You could try this with $N = \operatorname{Ker} f$. What is the relationship between $\operatorname{Ker} f$ and $\operatorname{Ker} f_\mathfrak{p}$?]

I should mention that if $M'$ is finitely presented then there is a relationship between $A$-linear $M' \to M$ and $S^{-1}A$-linear $S^{-1}M' \to S^{-1}M$. Proposition 2.10 of Eisenbud's book shows that in this situation we have a natural isomorphism $$ \operatorname{Hom}_{S^{-1}A}(S^{-1}M', S^{-1}M) \approx S^{-1}\operatorname{Hom}_A(M', M). $$ So you might suggestively call Georges' example "the identity map divided by $2$".

share|cite|improve this answer
What if i start with a sequence $0 \rightarrow M_{p,1} \rightarrow M_{p,2} \rightarrow M_{p,3} \rightarrow 0$? Can i get a sequence $M_1 \rightarrow M_2 \rightarrow M_3$? Note that at this point i only want to get the sequence, i am not worrying about exactness. – Manos Oct 31 '11 at 3:00
@Manos I've no idea (though I think it would be best to just talk about one homomorphism $M'_\mathfrak{p} \to M_\mathfrak{p}$ then, since that doesn't seem any easier/harder). I'm most worried about the case when, say, $M$ doesn't inject into $M_\mathfrak{p}$. So I'd like to play around with examples like that and try to think about the problem geometrically. But your best bet is probably Georges :) – Dylan Moreland Oct 31 '11 at 3:31
Dear Manos, you have had two answers to your original question. If this does not satisfy you, I suggest you write another, precise question with quantificators in front of $p$. In particular the example in my post shows that the answer to your question above is "no" since if you cannot lift $0 \to M_p \to N_p \to 0$, you certainly cannot lift $0 \to M_p \to N_p \to 0_p \to 0$ – Georges Elencwajg Oct 31 '11 at 10:03
@Dylan: thanks Dylan, you revealed the point that was confusing me: i already have $f$, no need to do any lifting. – Manos Oct 31 '11 at 13:46
Since no one has mentioned it, perhaps it is worth noting that the functor $M\rightarrow M_p$ is the tensor product $\otimes_A A_p$. So we know certain things about it on more-general principles... – paul garrett Oct 31 '11 at 13:48

In general, given an $A_p$-linear morphism $f:M_p\to N_p$ it is impossible to find a morphism of $A$-modules $g: M\to N$ lifting $f$ in the sense that $g_p=f.$

For example, let $A,M$ and $N$ all be equal to $\mathbb Z$. Choose for $p$ the zero ideal $p=(0)$.
Then $A_p=M_p=N_p=\mathbb Q$ and you can choose for $f:\mathbb Q \to \mathbb Q$ the morphism $f(q)=\frac{q}{2}$.
The only $\mathbb Z$-linear maps $g:\mathbb Z \to \mathbb Z$ are of the form $g(n)=an$ for some $a\in \mathbb Z$ and clearly none of them lifts $f$.

share|cite|improve this answer
The issue is that the exercise 10b) says to prove that the sequence $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$ is exact if and only if the sequence $0 \rightarrow M_{p,1} \rightarrow M_{p,2} \rightarrow M_{p,3} \rightarrow 0$ is exact. The forward direction is not a problem. The backward direction however requires some kind of lifting of the sequence to the initial modules. – Manos Oct 31 '11 at 0:27
@Manos: Take A=Z and p=0 as Georges suggests, then take M1 = 0, M2=Z/2Z, M3=0. Then the original sequence is not exact, but it becomes exact after localizing at p (it becomes 0→0→0→0→0). Presumably you need some extra hypotheses. – Jack Schmidt Oct 31 '11 at 1:23
@Manos You really want to require that the second sequence is exact for all primes $\mathfrak{p}$ of $A$. – Dylan Moreland Oct 31 '11 at 1:32
@Dylan: you are right, this is it. However, the question of how to lift back the sequence still remains. – Manos Oct 31 '11 at 1:59
@Manos Oh, do you want conditions under which a map $M_\mathfrak{p} \to N_\mathfrak{p}$ comes from an $M \to N$? That seems interesting. – Dylan Moreland Oct 31 '11 at 2:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.